Answer:
(x, y) = (2, 5)
Step-by-step explanation:
I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...
3x' -y' = 13/10
x' +2y' = 9/10
Adding twice the first equation to the second, we get ...
2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)
7x' = 35/10 . . . . . . simplify
x' = 5/10 = 1/2 . . . . divide by 7
Using the first equation to find y', we have ...
y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5
So, the solution is ...
x = 1/x' = 1/(1/2) = 2
y = 1/y' = 1/(1/5) = 5
(x, y) = (2, 5)
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The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.
Answer:
C
Step-by-step explanation:
The answer is D, because you move to the right of the decimal to see if that number is 5 or above. Then you round up, (to 86). :) Hope this was helpful.
Answer:
10 and 15
Step-by-step explanation:
Let 'x' and 'y' are the numbers we need to find.
x + y = 25 (two numbers whose sum is 25)
(1/x) + (1/y) = 1/6 (the sum of whose reciprocals is 1/6)
The solutions of the this system of equations are the numbers we need to find.
x = 25 - y
1/(25 - y) + 1/y = 1/6 multiply both sides by 6(25-y)y
6y + 6(25-y) = (25-y)y
6y + 150 - 6y = 25y - (y^2)
y^2 - 25y + 150 = 0 quadratic equation has 2 solutions
y1 = 15
y2 = 10
Thus we have
:
First solution: for y = 15, x = 25 - 15 = 10
Second solution: for y = 10, x = 25 - 10 = 15
The first and the second solution are in fact the same one solution we are looking for: the two numbers are 10 and 15 (since the combination 10 and 15 is the same as 15 and 10).
Answer:
Assuming that Janet spent an equal amount of time on each question, he would have spent 75 seconds on each question.
Step-by-step explanation:
10 minutes = 600 seconds
600 seconds / 8 questions = 75 seconds.
